20x^2+40=24x^2-40

Solution for 20x^2+40=24x^2-40 equation:

20x^2+40=24x^2-40

We move all terms to the left:

20x^2+40-(24x^2-40)=0

We get rid of parentheses

20x^2-24x^2+40+40=0

We add all the numbers together, and all the variables

-4x^2+80=0

a = -4; b = 0; c = +80;
Δ = b2-4ac
Δ = 02-4·(-4)·80
Δ = 1280
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1280}=\sqrt{256*5}=\sqrt{256}*\sqrt{5}=16\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{5}}{2*-4}=\frac{0-16\sqrt{5}}{-8}
=-\frac{16\sqrt{5}}{-8}
=-\frac{2\sqrt{5}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{5}}{2*-4}=\frac{0+16\sqrt{5}}{-8}
=\frac{16\sqrt{5}}{-8}
=\frac{2\sqrt{5}}{-1}
$